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Six Bugs (Posted on 2004-03-16) Difficulty: 4 of 5
Remember "Four Bugs" or "Three Bugs"?

In this problem, the six bugs start at the corners of a regular hexagon (with side length=10 inches).

Again, the bugs travel directly towards their neighbor (counter-clockwise). And, again, each bug homes in on its target, regardless of its target's motion. So, their paths will be curves spiraling toward the center of the hexagon, where they will meet.

What distance will the bugs have covered by then, and how did you determine it?

See The Solution Submitted by SilverKnight    
Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Where's Penny | Comment 3 of 8 |
Penny must be on spring break or she probably would have already commented that she posted the solution to the N-bug problem for general N as the first comment to "Three Bugs" http://perplexus.info/show.php?pid=1591&cid=11229 .
The solution she gives there is

"Then each bug will have gone:

V*T = L/(1-cos[2*pi/N radians]) "

For L=10, N=6

10/(1-cos[2*pi/6 radians])
= 10/(1-cos[60 degrees])
= 10/0.5
=20

  Posted by Richard on 2004-03-21 12:22:04
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